Question

给出一个列表，例如。 [2,3,4,10,20,30,102] 我需要所有可能的长度为3的块/子列表，如下所示，

[[2, 3, 4], [3, 4, 10], [4, 10, 20], [10, 20, 30], [20, 30, 102]]

Answer 1

这将完成这项工作：

l= [2,3,4,10,20,30,102] 
res=[l[i:i+3] for i in range(len(l)-2)]
print(res)

打印

[[2, 3, 4], [3, 4, 10], [4, 10, 20], [10, 20, 30], [20, 30, 102]]

Answer 2

这个功能可以完成更多的工作

def chunks(sequence, length):
    sub_sequences = [sequence[offset:] 
                     for offset in range(length)]
    return zip(*sub_sequences)

为你的例子

list(chunks([2, 3, 4, 10, 20, 30, 102], length=3))

提供所需的输出

Answer 3

使用list comprehension和xrange操作会帮到您。

示例输出

>>> a = [2,3,4,10,20,30,102]

>>> max_len = 3
>>> [ a[i-max_len: i] for i in xrange(max_len, len(a))]
[[2, 3, 4], [3, 4, 10], [4, 10, 20], [10, 20, 30]]

>>> max_len = 5
>>> [ a[i-max_len: i] for i in xrange(max_len, len(a))]
[[2, 3, 4, 10, 20], [3, 4, 10, 20, 30]]

Answer 4

arr = [2,3,4,10,20,30,102]
arr2 = arr
arr3 =[]
while (len(arr2) >= 3):
   arr4=arr2[:3]
   arr2 = arr2[1:]
   arr3.append(arr4)
print("Original List \n")
print(arr)
print("\n")
print("List with all the chunks \n")
print(arr3)

Answer 5

最简单的方法是使用切片：

>>> s = [2, 3, 4, 10, 20, 30, 102]
>>> [s[i:i+3] for i in range(0, len(s)-2)]
[[2, 3, 4], [3, 4, 10], [4, 10, 20], [10, 20, 30], [20, 30, 102]]

python中所有可能的列表块

5 个答案: